Question

Two particles each with a charge -Q are fixed a distance L apart . Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles. As a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is:_________.
A. Q
B. Q/2
C. Q/4
D. Q/8
E. Q/16

Answer 1

Answer:

B. Q/2

Explanation:

Given;

charge of the two particles, = -Q

distance between the two charges, r = L

the net force experienced by each charge = F

Apply coulomb's force between two charged particles;

$F = \frac{KQ^2}{r^2} = \frac{kQ^2}{L^2}$

when

+q is introduced midway between the charges, the net force is reduced by 2

$\frac{F}{2} = \frac{kQq}{L^2} \\\\ \frac{1}{2}(\frac{kQ^2}{L^2} ) = \frac{kQq}{L^2}\\\\\frac{kQ^2}{2L^2} =\frac{kQq}{L^2} \\\\\frac{Q}{2} = q$

Thus, The value of q is Q/2

Answer 2

The value of q is Q/2, when the net electric force experienced by each negatively charged particle is reduced to F/2.

Given here,  

-Q = charge of the two particles  

r = distance between the two charges = L  

F = net force on each charge  

From coulomb's force between two charged particles;

$\bold{F = \dfrac {KQ^2}{r^2} =\dfrac {kQ^2}{L^2} }$

If +q is bring at mid point between the charges, the net force is reduced by 2

Thus, q = Q/2

Therefore,   the value of q is Q/2, when the net electric force experienced by each negatively charged particle is reduced to F/2.